How should I solve $\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\ dx$?

Question

This integral $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\ dx$$

appears when I am trying to find the surface area of the solid of revolution generated by rotating $y=\frac{1}{x}$, $x\ge 1$ about $x$-axis.

I know that I can use $g\le f$ on $$[a,\ b]\implies \int_a^b g(x)\ dx\le\int_a^b f(x)\ dx$$ to prove the surface area approaches infinity. But, I still want to know how to integrate this by hand.

Answer 1

Perform $u$ substitution with $u=\sqrt{1+1/x^4}$ ;

To get $\int 1/x \sqrt{1+1/x^4}dx=-1/2\int u^2/(u^2-1)du=-1/2$$\int 1/(u^2-1) +1du$

To solve $\int 1/(u^2-1)du$ use difference of squares then partial fractions and the fact that $\ln(u)+c=\int 1/u du$

Answer 2

$$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\ dx$$ $$t = \sqrt{1 + \frac{1}{x^4}}$$ $$\frac{dt}{dx}=\frac{\frac{-4}{x^5}}{2\sqrt{1 + \frac{1}{x^4}}}$$ $$dx=\frac{2\sqrt{1 + \frac{1}{x^4}}}{\frac{-4}{x^5}}dt$$ $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\ dx=\frac{-1}{2}\int 2t^2 *x^4dt$$ $$=\frac{-1}{2}\int \frac{2t^2}{t^2-1}dt$$ $$=\frac{-1}{2}\int \Bigl(2+\frac{2}{t^2-1}\Bigr)dt$$ $$=-t -\frac{1}{2}\int \Bigl(\frac{2}{t^2-1}\Bigr)dt$$ $$=-t -\int \Bigl(\frac{1}{t^2-1}\Bigr)dt$$ $$=-t -\frac{1}{2}\ln\Bigl({\frac{t-1}{t+1}}\Bigr)+c$$ $$=-\sqrt{1 + \frac{1}{x^4}} -\frac{1}{2}\ln\Bigl({\frac{\sqrt{1+x^4}-x^2}{\sqrt{1+x^4}+x^2}}\Bigr)+c$$

Answer 3

Hint:

Let $x=t^{-a}$.

$$\int\frac1x{\sqrt{1+\frac1{x^4}}}\,dx=-a\int t^{-1}\sqrt{1+t^{4a}}\,dt.$$

It seems to make sense to choose one of $a=\frac12,a=\frac14$.

We take $a=\frac12$ and with $t=\sinh u$,

$$\int\frac{\sqrt{1+t^2}}t\,dt=\int \frac{\cosh u}{\sinh u}\cosh u\,du=\int \left(\sinh u+\frac{\sinh u}{\cosh^2 u-1}\right)\,du$$

which is immediate.

$$\cosh u+\text{artanh}(\cosh u)=\sqrt{1+t^2}+\text{artanh}(\sqrt{1+t^2})\\=\sqrt{1+x^{-4}}+\text{artanh}(\sqrt{1+x^{-4}}).$$